(log x)^ 2+(log y)^2-log(xy)log(x/y)=
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nice question
Answer:
ANSWER: NO! 2log (x + y) = log x + log y – 2log 3 does not show that x² + y² = 7xy; It shows instead that x² + y² = –17xy/9.
PROOF:
By the “logarithm of a product” property, we have:
2log(x + y) = log (xy) – 2log 3
Now, by the “logarithm of a power” property, we have:
log (x + y)² = log (xy) – log 3²
log (x + y)² = log (xy) – log 9
Now, by the “logarithm of a quotient” property, we have:
log (x + y)² = log (xy/9)
Now, let n = log (x + y)² and m = log (xy/9), then, substituting, we have the equality:
n = m
By the definition of logarithms and due to the fact that logarithms are exponents, the two logarithmic statements, n = log (x + y)² and m = log (xy/9), can be rewritten into their equivalent exponential forms as follows (We’re assuming that the logarithms that we’re dealing with in this problem are common logarithms (logarithms to the base 10):
10ⁿ = (x + y)² and 10ᵐ = xy/9.
Now, by a property of exponents which says that “if b > 0, b ≠ 1, and “m” and “n” are real numbers, then bⁿ = bᵐ if and only if n = m”; therefore, we have the following equality:
10ⁿ = 10ᵐ
(x + y)² = xy/9
x² + 2xy + y² = xy/9
9(x² + 2xy + y²) = (xy/9)9
9x² + 18xy +9y² = xy
9x² + 9y² = –18xy + xy
9x² + 9y² = –17xy
x² + y² = –17xy/9
CORRECTION NOTE: To get the desired result of x² + y² = 7xy, the original equation above has to have a plus (+) sign, rather than a minus (–) sign, before the last term on the right side, 2log 3, as follows:
2log (x + y) = log x + log y + 2log 3
PROOF:
Now, by the “logarithm of a product” property, we have:
2log(x + y) = log (xy) + 2log 3
Now, by the “logarithm of a power” property, we have:
log (x + y)² = log (xy) + log 3²
log (x + y)² = log (xy) + log 9
Again, by the “logarithm of a product” property, we have:
log (x + y)² = log (9xy)
Now, let n = log (x + y)² and m = log (9xy), then, substituting, we have the equality:
n = m
By the definition of logarithms and due to the fact that logarithms are exponents, the two logarithmic statements, n = log (x + y)² and m = log (9xy), can be rewritten into their equivalent exponential forms as follows (We’re assuming that the logarithms that we’re dealing with in this problem are common logarithms (logarithms to the base 10):
10ⁿ = (x + y)² and 10ᵐ = 9xy.
Now, by a property of exponents which says that “if b > 0, b ≠ 1, and “m” and “n” are real numbers, then bⁿ = bᵐ if and only if n = m”; therefore, we have the following equality:
10ⁿ = 10ᵐ
(x + y)² = 9xy
x² + 2xy + y² = 9xy
x² + y² = 9xy – 2xy which finally gives us the desired result:
x² + y² = 7xy